Finding the Limit: The Taylor Series Approach

[squeetox]

Homework Statement

I need to find the following limit.

Homework Equations

\lim_{x\rightarrow0}\frac{(x-\sinh x)(\cosh x- \cos x)}{(5+\sin x \ln x) \sin^3 x (e^{x^2}-1)}

The Attempt at a Solution

I think it's got to be something with Taylor series, but I don't really know how to do it.

 

[tiny-tim]

Welcome to PF!

\lim_{x\rightarrow0}\frac{(x-\sinh x)(\cosh x- \cos x)}{(5+\sin x \ln x) \sin^3 x (e^{x^2}-1)}
…
I think it's got to be something with Taylor series, but I don't really know how to do it.

Hi squeetox! Welcome to PF!

Hint: for example, the Taylor series for e^3x is 1 + 3x + … , so if (e^3x - 1) was a factor, you could replace it (as x -> 0) by 3x.

 

[squeetox]

Hi squeetox! Welcome to PF!

Thanks!


Why do the others terms (9x²/2 + 9x³/2...) disappear as x->0?

PS: I had thought of replacing \sin^3 x (e^{x^2}) \sim x^5 because I know that \sin [f(x)] \sim f(x) and e^{f(x)} -1 \sim f(x). The rest I have no clue though.

 

[tiny-tim]

Hi squeetox!

Why do the others terms (9x²/2 + 9x³/2...) disappear as x->0?

They don't actually disappear … but, compared with x, they get so small (as x -> 0) that we can ignore them.

PS: I had thought of replacing \sin^3 x (e^{x^2}) \sim x^5 because I know that \sin [f(x)] \sim f(x) and e^{f(x)} -1 \sim f(x). The rest I have no clue though.

_{Nooo … you're confusing e}x² _{and e}x² _{- 1.}

 

[springo]

They don't actually disappear … but, compared with x, they get so small (as x -> 0) that we can ignore them.

So basically you mean that since it's a sum of x's to different powers, and the ones after 3x get smaller much faster, I can ignore them?
Thanks.

 

[squeetox]

_{Nooo … you're confusing e}x² _{and e}x² _{- 1.}

I'm sorry that's what I meant. If you look at the limit I'm looking for, it's e^{x^2} - 1 \sim x^2 therefore sin^3 x(e^{x^2}-1) \sim x^5.

Now what about (x-\sinh x)(\cosh x- \cos x) and 5+\sin x \ln x?

 

[tiny-tim]

I'm sorry that's what I meant. If you look at the limit I'm looking for, it's e^{x^2} - 1 \sim x^2 therefore sin^3 x(e^{x^2}-1) \sim x^5.

ah yes, so it is … i didn't look back up to the top!

Now what about (x-\sinh x)(\cosh x- \cos x) and 5+\sin x \ln x?

The first lot, you do the same way.

The sinx lnx … I can't remember whether there's a Taylor expansion for lnx … it obviously does -> 0 … just look at sin10^-100 ln10^-100, but how to prove it? …

anyway, that's you problem, not mine!

 

[squeetox]

For sin x ln x, I'm thinking...
\sin x \ln x \sim x \ln x

And
\lim_{x\rightarrow0}x \ln x = \lim_{x\rightarrow\infty} \frac{1}{x}\ln\frac{1}{x} = \lim_{x\rightarrow\infty} -\frac{\ln x}{x} = 0

Therefore:
\lim_{x\rightarrow0}(5+\sin x \ln x) = 5


For the first lot, do I find Taylor expressions for (x - sinh x) and (cosh x - cos x) or for the product of both?

Edit: I got it:
x-\sinh x \sim -\frac{x^3}{6}
\cosh x - \cos x \sim x^2

So finally...
\lim_{x\rightarrow0}\frac{(x-\sinh x)(\cosh x- \cos x)}{(5+\sin x \ln x) \sin^3 x (e^{x^2}-1)} = -\frac{1}{30}

Thank you so much for your help!